2.

$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$

$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$

Vì: $0\leq \sin ^22x\leq 1$

$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$

Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$

3.

$0\leq |\sin x|\leq 1$

$\Rightarrow 3\geq 3-2|\sin x|\geq 1$

Vậy $y_{\min}=1; y_{\max}=3$

1.

\(y=\cos x+\cos (x-\frac{\pi}{3})=\cos x+\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(=\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(y^2=(\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)^2\leq (\cos ^2x+\sin ^2x)(\frac{9}{4}+\frac{3}{4})\)

\(\Leftrightarrow y^2\leq 3\Rightarrow -\sqrt{3}\leq y\leq \sqrt{3}\)

Vậy $y_{\min}=-\sqrt{3}; y_{max}=\sqrt{3}$

Chọn B

Đặt \(sinx+cosx=t\Rightarrow-\sqrt{2}\le t\le\sqrt{2}\)

\(t^2=sin^2x+cos^2x+2sinx.cosx=1+2sinx.cosx\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

\(\Rightarrow y=t+\dfrac{t^2-1}{2}=\dfrac{1}{2}t^2+t-\dfrac{1}{2}\)

Xét hàm \(f\left(t\right)=\dfrac{1}{2}t^2+t-\dfrac{1}{2}\) trên \(\left[-\sqrt{2};\sqrt{2}\right]\)

\(-\dfrac{b}{2a}=-1\) 

\(f\left(-\sqrt{2}\right)=\dfrac{1-2\sqrt{2}}{2}\) ; \(f\left(-1\right)=-1\) ; \(f\left(\sqrt{2}\right)=\dfrac{1+2\sqrt{2}}{2}\)

\(\Rightarrow y_{min}=-1\) khi \(t=-1\) ; \(y_{max}=\dfrac{1+2\sqrt{2}}{2}\) khi \(t=\sqrt{2}\)

Đặt \(t=sinx+cosx;t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow\dfrac{t^2-1}{2}=sinx.cosx\)

\(y=t+\dfrac{t^2-1}{2}=\dfrac{t^2}{2}+t-\dfrac{1}{2}\)

Vẽ BBT của \(f\left(t\right)=\dfrac{t^2}{2}+t-\dfrac{1}{2};t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow\)\(f\left(t\right)_{min}=-1\Leftrightarrow t=-1\Rightarrow sinx+cosx=-1\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt{2}}\)....

\(f\left(t\right)_{max}=\dfrac{1+2\sqrt{2}}{2}\)\(\Leftrightarrow t=\sqrt{2}\Rightarrow sinx+cosx=\sqrt{2}\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=1\)....

\(\Leftrightarrow m.cosx+\left(2m-1\right)sinx+3-m=y\left(cosx+sinx-2\right)\)

\(\Leftrightarrow\left(m-y\right)cosx+\left(2m-y-1\right)sinx=m-2y-3\)

Pt có nghiệm khi:

\(\left(m-y\right)^2+\left(2m-y-1\right)^2\ge\left(m-2y-3\right)^2\)

\(\Leftrightarrow2y^2+\left(2m+10\right)y-4m^2-2m+8\le0\)

\(\Rightarrow\dfrac{-m-5-\sqrt{9m^2+14m+9}}{2}\le y\le\dfrac{-m-5+\sqrt{9m^2+14m+9}}{2}\)

\(\Rightarrow y_{min}=\dfrac{-m-5-\sqrt{9m^2+14m+9}}{2}\le3\)

\(\Rightarrow\sqrt{9m^2+14m+9}\ge-m-11\)

BPT này đúng với mọi m. Vậy bài toán thỏa mãn với mọi m

Đáp án C

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